From: jonathan@iron.astro.indiana.edu   
      
   John Heath  wrote:   
   > 2 objects of equal mass are moving away from each other at .1 c   
   > caused by an explosion between them . Equal mass and equal speed so   
   > momentum is conserved. One fly in the soup.  One of the masses is   
   > radioactively decaying losing mass over time. How is momentum conserved   
   > under these conditions. I do not have a clear answer to this question.   
      
   The key point is that the *radioactive decay products can also carry   
   linear momentum*.   
      
      
      
   To look at this in detail, let's start by focusing on the mass (let's   
   say it's an atom A) which is radioactively decaying.  For simplicity   
   let's say that the decay is   
       A --> A' + e+ + e-   
   where the decay emits (only) a positron and an electron, which (let's   
   also assume) are emitted in opposite directions as seen from the atom.   
   Here A' is the post-decay atom.   
      
   It's instructive to first consider A's decay in the inertial reference   
   frame in which *A is at rest* (and hence has zero linear momentum) before   
   the decay.  Then by assumption the post-decay state has a positron (e+)   
   moving in some direction (let's call it the +y direction), and an electron   
   (e+) moving in the opposite direction (the -y direction).  Conservation   
   of momentum implies that the total linear momentum of A' + e- + e+ must   
   be the same as that of the pre-decay A, i.e., zero in this reference   
   frame.  Since the positron and electron have identical rest masses and   
   (we assumed) are moving in opposite directions (the +/- y directions),   
   their total linear momentum sums to zero (in this reference frame).   
   Hence A' must be at rest (in this reference frame) in order for the   
   total linear momentum of the 3 bodies A', e+, and e-, to sum to zero.   
   (= the total linear momentum before the decay).   
      
   Now let's consider the original system of two objects of equal mass,   
   moving apart, one of them (A) about to decay, the other (B) not decaying.   
      
   In the A-at-rest reference frame, the presence of B doesn't change our   
   description of A's decay -- B's linear momentum doesn't change during   
   A's decay, and the total momentum of   
     A + B  before decay   
   is the same as that of   
     (A' + e+ + e-) + B  after decay   
   (and is in fact precisely equal to the momentum of B alone, since   
   we've chosen a reference frame where A is at rest and thus has zero   
   momentum).   
      
   So everything is ok with conservation of momentum so far.   
      
      
      
   Now let's *change reference frames* to the one you probably thought of   
   first, in which A-before-decay and B have opposite velocities of +/- 0.1c   
   (let's say in the +x and -x directions respectively).  Since A-before-decay   
   and B are of equal mass, this means that the total linear momentum in   
   this reference frame is zero.  We call this the "AB-center-of-mass"   
   reference frame.   
      
   The key point is that in this reference frame A-before-decay is moving   
   at 0.1c in the +x direction, and nothing in A's decay changes the x   
   velocity component.  Therefore, the positron and electron which A emits   
   during the decay will *also* have velocity vectors with +0.1c components   
   in the +x direction.  Since the positron and electron also are emitted   
   with some velocities in the +/- y directions, this means that they're   
   actually moving in some +x+y ("northeast) and +x-y ("southeast") directions   
   (in the AB-center-of-mass reference frame).   
      
   In other words, *in the AB-center-of-mass reference frame, because   
   A-before-decay was moving in the +x direction, A's radioactive decay   
   products also carry some net linear momentum in the +x direction*, just   
   enough to "balance the books".   
      
   We earlier found that A's velocity in the X direction in't changed during   
   the decay, so (in the AB-center-of-mass reference frame) A' will still   
   be moving at 0.1c in the +x direction.  (But A' has a smaller rest mass   
   than A did.)   
      
   So (still in the AB-center-of-mass reference frame), the total momentum   
   works out like this:   
      
   before decay:   
     B: moving at 0.1c in the -x direction   
     A: moving at 0.1c in the +x direction   
     total momentum: zero   
      
   after decay:   
     B: moving at 0.1c in the -x direction   
     A': moving at 0.1c in the +x direction (but A' has less mass than A)   
     e+: moving in the +x+y ("northeast") direction,   
         with 0.1c +x component and some +y component   
     e-: moving in the +x-y ("southeast") direction,   
         with 0.1c +x component and some -y component   
     total momentum: x component = zero   
                     y component = zero   
      
   ciao,   
   --   
   -- "Jonathan Thornburg [remove -animal to reply]"    
      Dept of Astronomy & IUCSS, Indiana University, Bloomington, Indiana, USA   
      currently visiting Max-Plack-Institute fuer Gravitationsphysik   
                         (Albert-Einstein-Institut), Potsdam-Golm, Germany   
      "There was of course no way of knowing whether you were being watched   
       at any given moment.  How often, or on what system, the Thought Police   
       plugged in on any individual wire was guesswork.  It was even conceivable   
       that they watched everybody all the time."  -- George Orwell, "1984"   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)